Difference between revisions of "1982 AHSME Problems/Problem 8"
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== Solution == | == Solution == | ||
− | + | Since <math>\binom{n}{1}</math>, <math>\binom{n}{2}</math>, and <math>\binom{n}{3}</math> form an arithmetic progression, <math>\binom{n}{2} - \binom{n}{1} = \binom{n}{3} - \binom{n}{2}</math>. Therefore, <cmath>\frac{n!}{2!\cdot(n-2)!}-\frac{n!}{1!\cdot(n-1)!}=\frac{n!}{3!\cdot(n-3)!}-\frac{n!}{2!\cdot(n-2)!}.</cmath> | |
− | + | Simplifying these expressions yields <math>\frac{n(n-1)}{2}-n = \frac {n(n-1)(n-2)}{6}-\frac{n(n-1)}{2}</math>. Multiplying both sides by <math>6</math> and collecting all terms on one side yields <math>n^3 - 9n^2 + 14n = 0</math>, which factors to <math>n(n-7)(n-2)=0</math>. The solutions to this equation are <math>n \in \{0, 2, 7\}</math>, but <math>n > 3</math> so the only valid answer is <math>n = \boxed{(\mathbf{B})\ 7}</math>. | |
~ab2024 | ~ab2024 | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1982|num-b=7|num-a=9}} | ||
+ | |||
+ | {{MAA Notice}} |
Latest revision as of 22:38, 29 June 2025
Problem
By definition, and
, where
are positive integers and
.
If
form an arithmetic progression with
, then
equals
Solution
Since ,
, and
form an arithmetic progression,
. Therefore,
Simplifying these expressions yields . Multiplying both sides by
and collecting all terms on one side yields
, which factors to
. The solutions to this equation are
, but
so the only valid answer is
.
~ab2024
See Also
1982 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.